x[0,1] f(x)=sinx/1+x^2
isn't 1+x^2 arctan?
I wasn't sure how to approach this problem, I know sin/cos = tan, so would i use 1/2 x dx/1+x^2? I have no clue haha. Help please.
As written, the integrand is as follows:
$\displaystyle \frac{\sin(x)}{1}\, +\, x^2$
I have a feeling you mean one of the following...?
$\displaystyle \mbox{a) }\,\frac{\sin(x)}{1\, +\, x^2}$
$\displaystyle \mbox{b) }\,\sin\left(\frac{x}{1\, +\, x^2}\right)$
Please confirm or correct. Thank you!